Correct Answer – C::D
Let `veca, vecb and vecc` lie in the x-y plane.
Let `veca = hati, vecb = -(1)/(2) hati and vecc = -(1)/(2) hati – (sqrt3)/(2)hatj`.
Therefore,
`|vecp+ vecq + vecr| = |lamda veca + mu vecb + vvecc|`
`= | lamda hati + mu(-(1)/(2)hati + (sqrt3)/(2)hatj) + v(-(1)/(2)hati- (sqrt3)/(2)hatj)|`
`= |(lamda – (mu)/(2) – (v)/(2))hati + (sqrt3)/(2)(mu – v) hatj|`
` = sqrt((lamda – (mu)/(2) – (v)/(2))^(2) + (3)/(4)(mu-v)^(2))`
`= sqrt(lamda^(2)+ mu^(2) + v^(2) -lamda mu -lamda v – mu v)`
`” “=(1)/(sqrt2) sqrt((lamda-mu)^(2)+ (mu-v)^(2) +(v-lamda)^(2)) ge (1)/(sqrt2) sqrt(1+1+4) = sqrt3 `
Hence, `|vecp + vecq + vecr|` can take a value equal to `sqrt3` and 2.